GR | # Login | Register
  GR9277 #28
GREPhysics.NET Official Solution    Alternate Solutions
A system is known to be in the normalized state described by the wave function
where the $Y^m_l(\theta,\phi)$ are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m=3 is

  1. 0
  2. 1/15
  3. 1/6
  4. 1/3
  5. 13/15

Quantum Mechanics}Probability

One doesn't actually need to know much (if anything) about spherical harmonics to solve this problem. One needs only the relation P=\sum_i |\langle Y^3_i |\psi (\theta,\phi)\rangle|^2. Since the problem asks for states where m=3, and it gives the form of spherical harmonics employed as Y^m_l, one can eliminate the third term after the dot-product.

So, the given wave function \psi(\theta,\psi)=\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right) gets dot-product'ed like |\langle Y^3_i |\psi (\theta,\phi)\rangle|^2\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6\right)\right)\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right)\right)=\frac{25+1}{30}=\frac{13}{15}, as in choice (E).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
2012-04-19 19:55:26
One can also deduce that the particle is more likely than not to have m = 5 by looking at the normalized wave function. E is the only possibility that fits.NEC
2006-08-21 04:37:41
No dirac symbols not anything. The Possibility is c1^2 + c2^2 as ina a simple mathematical problem in our case 25/30+1/30=26/30
2009-09-09 22:12:22
agreed, the probability of finding a system in a given state is always just the sum of the squares of the coefficients in the wavefunction.

Post A Comment!
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...